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Problem A. 394. (February 2006)

A. 394. Let a1,a2,...,aN be nonnegative reals, not all 0. Prove that there exists a sequence 1=n0<n1<...<nk=N+1 of integers such that

n1an0+n2an1+...+nkank-1<3(a1+a2+...+aN).

(5 pont)

Deadline expired on March 16, 2006.


Solution. Define aN+2+aN+3=...=0 as well and look for an infinite sequence 1=n0<n1<... of integers for which

\sum_{i=1}^\infty n_ia_{n_{i-1}} < 3\sum_{n=1}^\infty a_n.

Define the function f:[1,\infty)\toR such that f(x)=f([x]).

The required sequence is constructed randomly. Take a random variable t\in[0,1] of uniform distribution. Set n0=1 and n_i=\left[2\cdot e^{i-1+t}\right] for i=1,2,.... Then

 E\left(\sum_{i=1}^\infty n_ia_{n_{i-1}}\right) \le
E(n_1)\cdot a_0 + E
\left(\sum_{i=2}^\infty2\cdot e^{i-1+t}f(2\cdot e^{i-2+t})\right)=

=
E(n_1)\cdot a_0 + \sum_{i=2}^\infty\int_0^1
\big(2\cdot e^{i-1+t}f(2\cdot e^{i-2+t})\big)dt=

= E(n_1)\cdot a_0 + \int_0^\infty 2\cdot e^{u+1}f(2\cdot e^u)du
= E(n_1)\cdot a_0 + e\int_2^\infty f(x)dx =

 = E(n_1)\cdot a_0 + e\sum_{n=2}^Na_n.

The value of n1 is 2,3,4 or 5 if t<\ln\frac32, \ln\frac32\le t<\ln\frac42, \ln\frac42\le t<\ln\frac52, or \ln\frac52\le t<1, respectively. Therefore

E(n_1)=
\left(\ln\frac32\right)\cdot2 +
\left(\ln\frac42-\ln\frac32\right)\cdot3 +
\left(\ln\frac52-\ln\frac42\right)\cdot4 +
\left(1-\ln\frac52\right)\cdot5 \approx 2,985<3

and

E\left(\sum_{i=1}^\infty n_ia_{n_{i-1}}\right)
< 3a_1 + e(a_2+\dots+a_N).

There always exists a sequence 1=n0<n1<...<nk=N+1 such that

n1an0+n2an1+...+nkank-1<3a1+e(a2+a3+...+aN).


Statistics:

3 students sent a solution.
5 points:Erdélyi Márton, Paulin Roland.
0 point:1 student.

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